Thread: Another probability question

Lets say a roulette table, the chance of the ball landing on red or black is 50/50. Is someone able to tell me what the odds are of the ball landing on red or black, 2 times in a row, 3 times in a row, 4 times in a row, 5 times in a row and 6 times in a row, and how those odds are works out?

0.5ⁿ.
That's for landing on the same colour n times in a row.
If it's 50/50 then the chance of that particular colour is 0.5. And then the next time, and the time after that. The ball has no memory.
So the first one is 0.5, the second one is 0.5 x 0.5, followed by the rest.
In fact that probability is the same for any sequence.
If you get RBRB it's exactly the same probability as BBBB.


Someone correct me (rapidly) if I'm wrong - it's early in the morning and I haven't eaten yet.

1/2^n

Originally Posted by Dywyddyr

0.5ⁿ.
That's for landing on the same colour n times in a row.
If it's 50/50 then the chance of that particular colour is 0.5. And then the next time, and the time after that. The ball has no memory.
So the first one is 0.5, the second one is 0.5 x 0.5, followed by the rest.
In fact that probability is the same for any sequence.
If you get RBRB it's exactly the same probability as BBBB.


Someone correct me (rapidly) if I'm wrong - it's early in the morning and I haven't eaten yet.

That is like saying your chances of winning the lottery are 1 in 49. Weetabix needed I think! :P

Originally Posted by leohopkins

That is like saying your chances of winning the lottery are 1 in 49. Weetabix needed I think! :P

Huh?
Applying the same mathematics doesn't give 1/49.
You have a 1/49 (I know bugger all about the lottery but assume there's 49 balls) chance of getting the first ball right.
Then you have a 1/48 chance of getting the next ball right. (There is some sort of "memory" here since the odds of getting each successive ball correct increase due to the fact that there are fewer possibilities with each one taken out).
So to even get the first TWO correct it's 0.02 x 0.02 (to two sig fig) = 0.004.
Please, show me where I went wrong.